Find the angles without using trigonometry

Question

In the figure, given that $JA=JB$, $\angle BAK=20^{\circ} $, $\angle KAJ=50^{\circ} $, $\angle KBA=40^{\circ} $, Find $\angle BJK$.

I succeed in solving it using trigonometry (quite tedious), but the answer turn out to be very neat and i believe there must be a very simple way to solve it using plane euclidean geometry. (pretty sure it exist but i tried many construction and still don’t work) I want to see the solution without trigonometry.

Any hints or solution is greatly appreciated. BTW, the angle i found it $10^{\circ} $ (neat right?)

Answer 1

Here is a solution not using trigonometry:

Assume $H$ is the midpoint of $AB$. Let's extend $BK$ such that it intersects $JH$ at $L$. Then, it is obvious that $\angle LAB=40^{\circ}$, so $\angle KAL=20^{\circ}$, which means $KA$ is the angle bisector of $\angle BAL$. Hence, by the angle bisector theorem, we have: $$\frac{LK}{KB}=\frac{AL}{AB}.$$

---

Note that this theorem can be proved without using trigonometry; for example take a look at this link.

---

On the other hand, consider the figure below, and observe that $\angle LAS=30^{\circ}$. Thus $LS=\frac{AL}{2}$.

Therefore, since $\triangle JLS$ and $\triangle JHA$ are similar, we have:

$$\frac{JL}{LS}=\frac{JA}{AH} \\ \implies \frac{JL}{JA}=\frac{AL}{AB} \implies \frac{JL}{JB}=\frac{AL}{AB}=\frac{LK}{KB},$$

which implies that $JK$ is the angle bisector of $\angle BJL$. So, $\angle BJK=10^{\circ}$.

We are done.

Answer 2

Hint:

Triangle is isosceles so:

$\angle CBA=\angle CAB=70^o\Rightarrow \angle ACB=40^o$

Draw the circumcircle of ABC and mark it's center as O.

Draw a diameter from A to meet the circle at D. Clearly AD bisect angle ACB. We have:

$\angle BCD=\frac {40}2=20^o$

Draw radius OE such that it intersect BD, clearly OE is perpendicular bisector of chord BD, that means it is parallel with BC, because angle CBD is opposite to diameter CD , so it is $90^o$. In this way E is mid point of arc BD and we have:

$OE||CB\Rightarrow \angle DOE=\angle BCE=20^o$

If the extension of CK meets the circle at E that would mean CE is bisector of angle BCD and so $\angle BCK= \frac{20}2=10$

Now we have to show the extension of CK meets the circle at E:

In triangle ABK:

$\angle AKB=120^o$

Connect A and C to E,Mark intersection of AB with CE as G.

$\angle ACE=30^0$

In triangle ACG we have:

$\angle AGC=180-[(\angle ACG=30) +(\angle CAG=70)]=80^o $

So in triangle ACK'( K' is the intersection of CE with AK) we have:

$\angle AK'G = 180-(20+80)=80^o$

Angle K'GA is exterior angle of triangle BGK' so we have:

$\angle BK'G=80-40=40^o\Rightarrow \angle BK'A=40+80=120^o=\angle BKA$

that is K is coincident on K' and as a result the extension of CK meets the circle at E.

Answer 3

The analytic approach is short and easy. The distance from K to the midline appears to equal the distance to BJ. Which means that JK bisects angle BJF.

Answer 4

The following solution uses the geometric constellation of a regular polygon, this is a good way to solve this problem and similar ones, since the solution does not only explain why we have such a "coincidence" of cevians in a triangle, but it also shows how to construct similar problems.

The first step is to realize the given isosceles triangle $\Delta JAB$ and the cevians $AK$, $BK$ inside of a regular polygon. The angles from the problem are multiples of $10^\circ$, so we should be fine with an $18$-gon. Here is the picture:

The $18$-gon has the vertices labelled $0,1,2,3,\dots$ - and in order to have an easy notation, the reflection of the point $k$ w.r.t. the diameter $09$ is denoted by $k'$, so we can use only digits, possibly with a prime decoration to address the vertices. The triangle $\Delta JAB$ is then realized as $\Delta 077'$, i.e. $J=0$, $A=7$, $B=7'$. We draw the (in the picture thick) cevians / chords $A5'=75'$ and $B3=7'3$, and $K$ corresponds to their intersection. Let $\Omega$ be its center (of symmetry). We will use the notation $(\Omega_{18})$ for the whole regular $18$-gon $0123\dots898'\dots3'2'1'$.

A remark on the quick visualization of measures of inscribed angles in the figure. The angle $\widehat{BAK}=\widehat{7'75'}$ is opposed to the arc $\overset\frown{7'5'}$, which has "two base arcs" of measure $20^\circ$ (delimited by the vertices of the $18$-gon), so its measure is $\frac 12\cdot 2\cdot 20^\circ=20^\circ$. The angle $\widehat{JAB}=\widehat{077'}$ is opposed to the arc $\overset\frown{7'0}$, which has "seven base arcs" of measure $20^\circ$ (delimited by the vertices of the $18$-gon), so its measure is $\frac 12\cdot 7\cdot 20^\circ=70^\circ$. And so on. Inscribed angles that delimit "$k$ base arcs" have measure $k\cdot 10^\circ$. A similar observation holds for angles like $\widehat{AKB}=\widehat{7K7'}$, we have to add the "base arcs" $\overset\frown{77'}+\overset\frown{5'3}$, so the measure is $(4+8)\cdot 10^\circ=120^\circ$. For this reason no angles are explicitly marked in the picture.

The problem asks for the placement of the third cevian $JK$, and we want to show that it passes through $8'$. To have an isolated statement:

Lemma: In the given regular $18$-gon $(\Omega_{18})$ the following lines are concurrent in the point $K$:

• the side bisector $s$ of the (parallel, congruent) segments $12$ and $7'8'$, which is an axis of symmetry for the following pairs of lines,
• the pair of lines $75'$, and $84'$, mirrored w.r.t. to the $s$-reflection,
• the pair of lines $08'$, and $37'$, mirrored w.r.t. to the $s$-reflection.

Corollary: $$\widehat{BJK}=\widehat{7'08'}=\bbox[yellow]{10^\circ}\ .$$

---

Note: At a first glance, it seems it's only an overhead feature to introduce $(\Omega_{18})$. Well, this would be the case if the solution would come "without using it", by just using construction related in a direct manner to the initial triangle, and only indirect to $(\Omega_{18})$. Then only searching the solution would be motivated by this overhead. And the presentation would get a structural touch. However, this is not case, the idea to proceed in such cases is to use geometric versions of the given $18$-gon, like translations, rotations, rescalings of it, and the specific symmetries of these figures put together to conclude. And usually, there are more then one relevant isometries of the plane bringing one regular polygon into the other one, combining such isometries gives geometrical insight. So let us start the proof.

---

Story and Proof of the Lemma: The triangle $\Delta BKA=\Delta 7'K7$ has the same angles as the triangle $\Delta 7'5'1'$, namely $40^\circ$, $120^\circ$, and $20^\circ$. So we can imagine a rotated, rescaled version of $(\Omega_{18})$ (which is centered in $\Omega$ and passes through $7'$), call it $(\Xi_{18})$ (which is centered in a point $\Xi$ and also passes through $7'$), obtained by a geometrical transformation $$g$$ of $\Omega$, such that the "solid piece" $\Delta 7'5'1'$ from $(\Omega_{18})$ is mapped to $(\Xi_{18})$. The rotational part is centered in $7'$, and its angle is $\widehat{5'7'K}= 80^\circ$, and the rescaling / homotety is a contraction centered in $K$ with factor $7'K:7'5'= 7'7:7'1'$. Since i need from $(\Xi_{18})$ only the regular $9$-gon with a vertex in $7'$, i will draw only this part, $(\Xi_9)$, to plot less points. Here is the picture.

As a digression, note that there are more geometrical transformations from $(\Omega_{18})$ to $(\Xi_{18}$, for instance by rotation + homothety centered in $7'$ (as already mentioned), by rotation + homothety centered in $7$ (the $s$-reflected version), by a translation moving $\Omega\to \Xi$ followed by a $\Xi$-homothety,

Let $K'$ be the point such that $g$ maps the isosceles trapezium $7'5'3'1'$ into the rotated, similar trapezium $7'KK'7$. We obtain gratis the information $$ 7'K=KK'=K'7\ , \tag{$*$} $$ and $K'$ is the reflection of $K$ w.r.t. the symmetry axis $s=09$, the side bisector of $7'7$.

Let $S$ be the mid point of $KK'$, $S=KK'\cap 0\Omega9\Xi$, and let $T$ be the intersection $AK5'\cap JB$. Then $$ KS=\frac 12KK'=\frac 12KB=KT\ , $$ the last equality happens because there is a $90^\circ$ angle in $T$ in $\Delta BTK$, and a $30^\circ$ angle in $B$. So $K$ has equal distance to the lines $J\Omega 9$ and $JB$, so it is on the angle bisector $J8'$.

$\square$

---

---

We are done, but it is maybe good to notice further properties in the given picture. Please ignore, if this feels off topic.

Since $K$ is on $08'$ and $7'3$, it is also on the symmetry axis of the two lines, which is the side bisector of $12$ and $7'8'$. In particular $K7'=K8'$, so we can continue $(*)$ with one more segment length. The points $7',8',K'$ are thus on one circle centered in $K$, and we can further add special points in the list. Among them $\Omega$ is the striking one, this happens because $\Delta \Omega KK'$ is equilateral. (Proof: It is isosceles, and $\widehat{\Omega KK'}=\widehat{17'7}=60^\circ$.

Bonus: We can compute the angle $\widehat{K0K'}=\widehat{8'08}=20^\circ$. So in $(\Omega_{18})$ the isosceles triangle $\Delta 088'$ has the same shape as the reflection of $\Delta 0KK'$ w.r.t. $K'$. This explains the picture: